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Digiron [165]
3 years ago
15

On Monday, Melodie ate 14 snickers

Mathematics
1 answer:
Gelneren [198K]3 years ago
3 0

Answer:

3

Step-by-step explanation:

1 1/4 + 1 3/4 = 3. Hope this helps!

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A rectangular piece of paper has an area of 117 square inches and a perimeter of 44 inches. What are the dimensions of the paper
-Dominant- [34]

Answer:

13x9

Step-by-step explanation:

half of perimeret of  rectangle L+W

half of perimeter of  rectangle = 44/2 = 22

L+W=22

LxW=117

  Think of 2 numbers that added =22 and multiplied =117

       13+9 =22

    13x9 =117

3 0
2 years ago
Solve the equation 4y + 12 = 0.2 (20 + 60y)
Rudik [331]

Answer:

y=1

Step-by-step explanation:

8 0
3 years ago
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Simplify the expression <br> 2x (x - 6)
Ulleksa [173]
I’m pretty sure it’s 2x^2 - 12x
7 0
3 years ago
Lionel's front yard is in the shape of a square and has an area of 225 square meters
photoshop1234 [79]
One side of the square is 15 cause 15 times 15=225
3 0
3 years ago
Read 2 more answers
Can someone help me solve this?
nikdorinn [45]
So...the tent, seems to be the first picture below

the second picture, is just a break-down of the pyramid

\bf \textit{area of a square}=A=a^2=4.9\implies a=\sqrt{4.9}&#10;\\\\\\&#10;\textit{slant height will be}=c^2=\left( \frac{\sqrt{4.9}}{2} \right)^2+1^2&#10;\\\\\\&#10;c^2=\cfrac{4.9}{4}+1\implies c^2=\cfrac{89}{40}\implies c=\sqrt{\cfrac{89}{40}}

so... that's the value of "a" and the "c" or "slant height", namely the missing slanted side on that second picture

we need the slant height, in order to get the area of that triangular face, and the base of that triangle is, just "a"

since the area of a triangle is 1/2 bh, and the base is "a" and the altitude or height here is the slant height "c", then the area of that triangular face is \bf \left( \cfrac{1}{2} \right)\left( \sqrt{4.9}\right)\left( \sqrt{\cfrac{89}{40}} \right)

and the area of the lateral sides for the prism, are " a * 1.5"  or \bf \sqrt{4.9} \cdot 1.5

so, you have 4 triangles, and 4 rectangles to be added, so.. .let's do that, to get the canvas then

\bf \begin{array}{clclll}&#10;4\left[ \left( \cfrac{1}{2} \right)\left( \sqrt{4.9}\right)\left( \sqrt{\cfrac{89}{40}} \right) \right]&+&4[(\sqrt{4.9})(1.5)]\\&#10;\uparrow &&\uparrow \\&#10;\textit{4 triangles}&&\textit{4 rectangles}&#10;\end{array}

that'd be the area of the canvas, or namely, the surface area

notice, we are not including the base of the pyramid, because that's inside the tent, and excluding the base of the prism, because that'd be the ground, and the tent may not include that, well, depends on the tent, in this case, I think a tent this big may not, other smaller ones do though




5 0
3 years ago
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